given $x$ irrational can you find $a,b in mathbb{Q}$ such that $a+bx = r$ for all $r in mathbb{R}$












0














given $x$ irrational can you find $a,b in mathbb{Q}$ such that $a+bx = r$ for all $r in mathbb{R}$.



I'm trying to solve this. My attempt consists of choosing $b$ close enough to $bx$ such that $bx to 0$ and $a$ close enough to $r$ such that $a to r$.



Or in a rigorous sense:



choose $b in mathbb{Q}$ such that $bx = epsilon$



choose $a in mathbb{Q}$ such that $a = r - epsilon$



I can choose such $a,b in mathbb{Q}$ since $Q$ is dense in $mathbb{R}$






real-analysis abstract-algebra







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  • The title does not make sense. Given $x$, $a+bx$ is a fixed number for whatever $a,b$ one chooses. Do you mean that given $x$ irrational, and $rinmathbb{R}$, can one find $a,bin{mathbb Q}$ such that $a+bx=r$? Or you simply want to know if it is true that $mathbb{R}={a+bxmid a,binmathbb{Q}}$ for fixed irrational $x$?
    – user587192
    Dec 10 '18 at 13:52








  • 1




    You can't do it for all $rin Bbb R$, as there is no choice for $a, b$ which works simultaneously for $r = 2$ and at the same time for $r = 3$. For any $rin Bbb R$, on the other hand, is a different question, and probably the one you meant to ask.
    – Arthur
    Dec 10 '18 at 13:55


















0














given $x$ irrational can you find $a,b in mathbb{Q}$ such that $a+bx = r$ for all $r in mathbb{R}$.



I'm trying to solve this. My attempt consists of choosing $b$ close enough to $bx$ such that $bx to 0$ and $a$ close enough to $r$ such that $a to r$.



Or in a rigorous sense:



choose $b in mathbb{Q}$ such that $bx = epsilon$



choose $a in mathbb{Q}$ such that $a = r - epsilon$



I can choose such $a,b in mathbb{Q}$ since $Q$ is dense in $mathbb{R}$






real-analysis abstract-algebra







share|cite|improve this question






















  • The title does not make sense. Given $x$, $a+bx$ is a fixed number for whatever $a,b$ one chooses. Do you mean that given $x$ irrational, and $rinmathbb{R}$, can one find $a,bin{mathbb Q}$ such that $a+bx=r$? Or you simply want to know if it is true that $mathbb{R}={a+bxmid a,binmathbb{Q}}$ for fixed irrational $x$?
    – user587192
    Dec 10 '18 at 13:52








  • 1




    You can't do it for all $rin Bbb R$, as there is no choice for $a, b$ which works simultaneously for $r = 2$ and at the same time for $r = 3$. For any $rin Bbb R$, on the other hand, is a different question, and probably the one you meant to ask.
    – Arthur
    Dec 10 '18 at 13:55
















0












0








0







given $x$ irrational can you find $a,b in mathbb{Q}$ such that $a+bx = r$ for all $r in mathbb{R}$.



I'm trying to solve this. My attempt consists of choosing $b$ close enough to $bx$ such that $bx to 0$ and $a$ close enough to $r$ such that $a to r$.



Or in a rigorous sense:



choose $b in mathbb{Q}$ such that $bx = epsilon$



choose $a in mathbb{Q}$ such that $a = r - epsilon$



I can choose such $a,b in mathbb{Q}$ since $Q$ is dense in $mathbb{R}$






real-analysis abstract-algebra







share|cite|improve this question













given $x$ irrational can you find $a,b in mathbb{Q}$ such that $a+bx = r$ for all $r in mathbb{R}$.



I'm trying to solve this. My attempt consists of choosing $b$ close enough to $bx$ such that $bx to 0$ and $a$ close enough to $r$ such that $a to r$.



Or in a rigorous sense:



choose $b in mathbb{Q}$ such that $bx = epsilon$



choose $a in mathbb{Q}$ such that $a = r - epsilon$



I can choose such $a,b in mathbb{Q}$ since $Q$ is dense in $mathbb{R}$






real-analysis abstract-algebra




real-analysis abstract-algebra






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Dec 10 '18 at 13:48









Xenidia

1,275629




1,275629












  • The title does not make sense. Given $x$, $a+bx$ is a fixed number for whatever $a,b$ one chooses. Do you mean that given $x$ irrational, and $rinmathbb{R}$, can one find $a,bin{mathbb Q}$ such that $a+bx=r$? Or you simply want to know if it is true that $mathbb{R}={a+bxmid a,binmathbb{Q}}$ for fixed irrational $x$?
    – user587192
    Dec 10 '18 at 13:52








  • 1




    You can't do it for all $rin Bbb R$, as there is no choice for $a, b$ which works simultaneously for $r = 2$ and at the same time for $r = 3$. For any $rin Bbb R$, on the other hand, is a different question, and probably the one you meant to ask.
    – Arthur
    Dec 10 '18 at 13:55




















  • The title does not make sense. Given $x$, $a+bx$ is a fixed number for whatever $a,b$ one chooses. Do you mean that given $x$ irrational, and $rinmathbb{R}$, can one find $a,bin{mathbb Q}$ such that $a+bx=r$? Or you simply want to know if it is true that $mathbb{R}={a+bxmid a,binmathbb{Q}}$ for fixed irrational $x$?
    – user587192
    Dec 10 '18 at 13:52








  • 1




    You can't do it for all $rin Bbb R$, as there is no choice for $a, b$ which works simultaneously for $r = 2$ and at the same time for $r = 3$. For any $rin Bbb R$, on the other hand, is a different question, and probably the one you meant to ask.
    – Arthur
    Dec 10 '18 at 13:55


















The title does not make sense. Given $x$, $a+bx$ is a fixed number for whatever $a,b$ one chooses. Do you mean that given $x$ irrational, and $rinmathbb{R}$, can one find $a,bin{mathbb Q}$ such that $a+bx=r$? Or you simply want to know if it is true that $mathbb{R}={a+bxmid a,binmathbb{Q}}$ for fixed irrational $x$?
– user587192
Dec 10 '18 at 13:52






The title does not make sense. Given $x$, $a+bx$ is a fixed number for whatever $a,b$ one chooses. Do you mean that given $x$ irrational, and $rinmathbb{R}$, can one find $a,bin{mathbb Q}$ such that $a+bx=r$? Or you simply want to know if it is true that $mathbb{R}={a+bxmid a,binmathbb{Q}}$ for fixed irrational $x$?
– user587192
Dec 10 '18 at 13:52






1




1




You can't do it for all $rin Bbb R$, as there is no choice for $a, b$ which works simultaneously for $r = 2$ and at the same time for $r = 3$. For any $rin Bbb R$, on the other hand, is a different question, and probably the one you meant to ask.
– Arthur
Dec 10 '18 at 13:55






You can't do it for all $rin Bbb R$, as there is no choice for $a, b$ which works simultaneously for $r = 2$ and at the same time for $r = 3$. For any $rin Bbb R$, on the other hand, is a different question, and probably the one you meant to ask.
– Arthur
Dec 10 '18 at 13:55












2 Answers
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So you're asking if the simple field extension ${Bbb Q}(x) = {a+bxmid a,bin{Bbb Q}}$ can be equal to the real numbers? If so, note that the first set is denumerable, while the latter is not.






share|cite|improve this answer





























    2














    That cannot work out for cardinality reasons, if $x$ is fixed. lets assume you would be able to do that, then this would mean that you would get a surjective map
    $$mathbb{Q} times mathbb{Q} to mathbb{R} \ a , b mapsto a+bx.$$
    But the left set is countable, and the right one is uncountable, which produces a contradiction.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      6














      So you're asking if the simple field extension ${Bbb Q}(x) = {a+bxmid a,bin{Bbb Q}}$ can be equal to the real numbers? If so, note that the first set is denumerable, while the latter is not.






      share|cite|improve this answer


























        6














        So you're asking if the simple field extension ${Bbb Q}(x) = {a+bxmid a,bin{Bbb Q}}$ can be equal to the real numbers? If so, note that the first set is denumerable, while the latter is not.






        share|cite|improve this answer
























          6












          6








          6






          So you're asking if the simple field extension ${Bbb Q}(x) = {a+bxmid a,bin{Bbb Q}}$ can be equal to the real numbers? If so, note that the first set is denumerable, while the latter is not.






          share|cite|improve this answer












          So you're asking if the simple field extension ${Bbb Q}(x) = {a+bxmid a,bin{Bbb Q}}$ can be equal to the real numbers? If so, note that the first set is denumerable, while the latter is not.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 13:51









          Wuestenfux

          3,5791411




          3,5791411























              2














              That cannot work out for cardinality reasons, if $x$ is fixed. lets assume you would be able to do that, then this would mean that you would get a surjective map
              $$mathbb{Q} times mathbb{Q} to mathbb{R} \ a , b mapsto a+bx.$$
              But the left set is countable, and the right one is uncountable, which produces a contradiction.






              share|cite|improve this answer


























                2














                That cannot work out for cardinality reasons, if $x$ is fixed. lets assume you would be able to do that, then this would mean that you would get a surjective map
                $$mathbb{Q} times mathbb{Q} to mathbb{R} \ a , b mapsto a+bx.$$
                But the left set is countable, and the right one is uncountable, which produces a contradiction.






                share|cite|improve this answer
























                  2












                  2








                  2






                  That cannot work out for cardinality reasons, if $x$ is fixed. lets assume you would be able to do that, then this would mean that you would get a surjective map
                  $$mathbb{Q} times mathbb{Q} to mathbb{R} \ a , b mapsto a+bx.$$
                  But the left set is countable, and the right one is uncountable, which produces a contradiction.






                  share|cite|improve this answer












                  That cannot work out for cardinality reasons, if $x$ is fixed. lets assume you would be able to do that, then this would mean that you would get a surjective map
                  $$mathbb{Q} times mathbb{Q} to mathbb{R} \ a , b mapsto a+bx.$$
                  But the left set is countable, and the right one is uncountable, which produces a contradiction.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 13:54









                  Enkidu

                  95618




                  95618






























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